2E4: SOLIDS & STRUCTURES

Lecture 10

Dr. Bidisha GhoshNotes: http://www.tcd.ie/civileng/Staff/Bidisha.Ghosh/Solids & Structures

Components of Stress

• Elementary Cube: A infinitesimally small cube (dimensions: dxxdyxdz) is considered to get an insider’s view on the stresses acting inside the material of the beam.

• Nine components of stress are acting. (A stress tensor can be defined.)

• For homogenous, isotropic material,

• Hence, three normal stress components and three shear stress components can be considered.

; ; ;xy yx xz zx yz zy

Stress Tensor

Typical cases of stress

STRESS IS DIRECTIONALUniaxial Stress: An cube element of the body when subjected only to normal stresses along one direction only, the one-dimensional state of stress is referred to as uniaxial tension or compression. (Everything that has been done so far)(or in other words only is acting on the elementary cube.

Biaxial Stress: An cube element of the body when subjected only to normal stresses along two perpendicular directions. In this case,

x

; y yx xx yE E E E

Typical Cases of Stresses (that are modelled)

Tri-axial Stress: An cube element of the body when subjected only to normal stresses, acting in mutually perpendicular directions it is said to be in a state of triaxial stress.

The absence of shearing stresses indicates that the preceding stresses are the principal stresses for the element.

Typical cases of stress

Pure Shear: In this case, the element is subjected to plane shearing stresses only.

The planes ab and cd are in pure shear.

Typical pure shear occurs over the cross sections and on longitudinal planes of a circular shaft subjected to torsion.

x y

Typical Cases of Stresses (that are modelled)

Plane Stress: An elementary cube of a body under stress when subjected only to biaxial stresses only on the x and y faces of the element, then the body is in plane stress.The stresses can be both normal stresses and/or shear stresses.

This set of equation represent a state called plane stress and also Hooke’s law for 2D stress.(Extend it to 3D!!!)

1=

1

yxx x y

y xy y x

z y x

xy xy

E E E

E E E

E

G

• Stress element can be defined as stress gauge. Similar to strain gauge it helps to measure what is happening inside the body under stress.

• We will be concentrating only on 2D stress element for simplicity (and sanity!)

• 2D stress elements are actually an infinitesimally small layer cut out of the elementary cube under plane stress conditions.

• This will help us to know what is going on with stresses inside!!

Stress Element

Consider an inclined section of a prismatic bar. For equilibrium,

Assume no shearing stress working on the inclined plane.

Let’s name the inclined plane pq.

The force can be resolved to a component perpendicular to the plane pq, called

And, parallel to the plane pq,

The area of the inclined plane pq,

Stresses on Inclined Planes

00 cos ;

cos

AA A A

P P

PcosF P

sinV P

Stresses on Inclined Planes

Normal stress on pq,

Shear Stress on pq,

2cosn x 1

sin 22 x

Stresses on Inclined PlanesSo, what is the maximum normal stress?

And maximum shear stress,

•The formulae are valid in tension (positive) and compression (negative).

, 0n x

452x

Shear Failure of Column

Shear vs. Normal Stress

For ,

,normal stressn

,shear stress2cos

1sin 2

2

n x

x

= 1MPa

theta (degree)

normal stress

shear stress

0 1.00 0.0015 0.93 0.2530 0.75 0.4345 0.50 0.5060 0.25 0.4375 0.07 0.2590 0.00 0.00

105 0.07 -0.25120 0.25 -0.43135 0.50 -0.50150 0.75 -0.43165 0.93 -0.25180 1.00 0.00

-0.20 0.00 0.20 0.40 0.60 0.80 1.00 1.20

-0.60

-0.40

-0.20

0.00

0.20

0.40

0.60

Shear vs. Normal Stress

On the circle, the highlighted point is the topmost point on the circle. From the right most point (1,0) if we move anticlockwise, then the points are located at 2 angle.

theta (degree)

normal stress

shear stress

0 1.00 0.0015 0.93 0.2530 0.75 0.4345 0.50 0.5060 0.25 0.4375 0.07 0.2590 0.00 0.00

105 0.07 -0.25120 0.25 -0.43135 0.50 -0.50150 0.75 -0.43165 0.93 -0.25180 1.00 0.00

0.00 0.50 1.00-0.60

-0.40

-0.20

0.00

0.20

0.40

0.60

Series1; 0.00

0.25

0.43

0.50

0.43

0.25

0.00

-0.25

-0.43

-0.50

-0.43

-0.25

0.00

This circle is called Circle of Stress

X-axis normal stress, Y-axis shear stress

1. Pt. A Surface pq normal to axis of bar,

2. Pt. O Surface parallel to the axis of bar,

3. To obtain a point in this circle corresponding to the stresses on a plane at an angle of inclination it is only necessary to measure in counter-clockwise direction from point A an arc subtending an angle

0; ; 0;n x

; 0; 0;2 n

2

2cos ; sin 22x

n x

Circle of Stress

Take a plane normal to pq, called mm

The sum of normal stresses acting on two perpendicular planes remain constant and shearing stresses acting on two perpendicular planes are equal but opposite in direction.

2 2 21

1

;2 2

cos cos ( ) sin2

sin 2 sin 2( ) cos sin 2 sin 22 2 2 2 2

n x x x

x x x x

Example1

Solution

Combined Stress

Boiler surface

The element of the wall is undergoing tension in x and y direction.

For x direction, the stress is 2 1

cos ; sin 22n x x

x

Combined Stress

Boiler surface

For y direction, the stress isFollowing the same steps as before,The angle between line of application of force to the normal on the inclined surface is

Hence, 2 2cos ( ) sin

2

sin 2( ) sin 222 2

n y y

yy

y

2

Combined Stress

Hence, the total stress,

To obtain stress components for point D same procedure as before.

2 2cos sin ;

1( )sin 2 ;

2

n x y

x y

Combined Stress

Example 2

Determine the shear stress and normal stress for an angle of inclination (f ) = 30o and 120 o if . Isolate an element of the material and show the stresses by arrow.

http://www.dept.aoe.vt.edu/~jing/MohrCircle.html

Plane Stress0;z zx xz

There is an longitudinal stress , a transverse stress , and a shear stress. We have worked so far with axial stresses in two directions, now add a shear stress component to it. Hence, on the inclined plane now the stresses are,

x y

xy

2 2cos sin sin 2 ;

( )sin 2cos2 ;

2

n x y xy

x yxy

Stress Transformation

So now add the part for shearing force,

Mohr’s Circle for Planes Stress

2 2cos sin sin 2 ;

( )sin 2cos2 ;

2

n x y xy

y xxy

Principal StressThe maximum and minimum normal stresses are called the principal stresses.On the x-axis of Mohr’s circle, the maximum and minimum x-ordinate values are the principal stress values. (the red dots on the circle)

In case of plane stress situation, the principal stresses are,

Maximum Shear Stress

From Mohr’s circle of plane stress, we can say, the maximum shear stress is at the top of the Mohr’s circle.

Which is equal to the radius of the circle,

Principal planes

• The angle of the shaded triangle is double the angle of the principal plane (with maximum normal stress).

• The principal planes are perpendicular to each other.

• The most important stress element is the one that is along the principal plane showing principal stresses.

How to construct a Mohr’s Circle

Extra 1: Plane Strain

Occurs when the material is subjected to two normal strains and one shear strain.(Similar to plane stress)

0;z zx xz

Extra 2: Strain Gauges

Extra 3: General Mohr’s Circle (too complex)

Extra 4